In the closing decades of the nineteenth century, the German chemist Emil Fischer set himself a problem that looked impossible. Everyone agreed that glucose had the molecular formula \(\mathrm{C_6H_{12}O_6}\). But the molecule carries four chiral carbon atoms, and a structure with four chiral centres can exist in sixteen different three-dimensional arrangements. Sixteen molecules, identical on paper, different in space. Which one was the glucose in grapes, in honey, in our own blood?
Fischer had no X-ray machine and no spectrometer. He had test tubes, patience, and logic. He oxidised sugars, reduced them, converted one sugar into another, and compared the optical rotations of the products. Each result ruled out some of the sixteen candidates and kept others alive. Over years of such work he narrowed the field down to one. Along the way he devised a simple flat way of drawing a three-dimensional molecule on paper — the Fischer projection formula that you will use again and again in this chapter. In 1902 he received the Nobel Prize in Chemistry for his work on sugars and purines.
This chapter is about that idea: the molecules of life are not just lists of atoms. Their exact shape decides what they can do — which sugar a plant stores, which one builds wood, why a boiled egg never turns back to liquid, and how a handful of small bases can carry the instructions for an entire organism.
Q1. Glucose, maltose and sucrose are all sugars, and sucrose is even built out of glucose. Why then do glucose and maltose give a positive Tollens test, while sucrose gives a negative one? Q2. Starch and cellulose are both long chains made only of glucose units. Why can we digest starch easily but not cellulose at all, even though the building block is the same? Q3. When an egg is boiled, no peptide bond in its protein is broken — yet the change is permanent and the egg white never becomes clear liquid again. What exactly has changed in the protein? Q4. Glucose is written as D-(+)-glucose and fructose as D-(-)-fructose. Both carry the letter D, but one sign is positive and the other negative. What does the letter tell us, and what does the sign tell us — and why are they not the same information? Q5. DNA is built from only four different bases, yet it stores the instructions for making thousands of different proteins in one organism. How can so few building blocks hold so much information?
Keep these questions in mind as you read. The chapter will give you everything you need to reason out each one for yourself.
14.1 Introduction: Principal Molecules of the Living World
- What are the constituents of a balanced diet?
- What are the products of digestion of carbohydrates?
- Which constituent of diet is useful for building muscles?
- Which constituent of diet is a source of high energy?
- What is the genetic material of organisms?
The body of any living organism — a bacterium, a mango tree, a human being — is built from a very large number of different molecules. Some of these molecules make up the physical structure of the body. Others take part in the physiological processes that keep the organism alive: digestion, respiration, growth, repair, and the passing on of characters to the next generation.
Biomolecules are the molecules that build the structure of living organisms and that take part in the physiological processes occurring inside them.
What This Chapter Covers
Of these four classes, this chapter takes up three in detail: carbohydrates, proteins and nucleic acids. For each one you will study how the molecule is built, how its structure is represented on paper, and how that structure explains what the molecule does inside a living body.
For a long time chemists believed that the substances found in living things could be made only inside a living body, by a mysterious “vital force” that no laboratory could copy. That belief was broken by the German chemist Friedrich Wöhler in 1828, when he heated ammonium cyanate — an ordinary inorganic salt — and obtained urea, a compound excreted by animals. A molecule of the living world had been made from a lifeless starting material. The study of biomolecules as ordinary chemistry, which this chapter is about, begins with that result.
The link between soil minerals and plant protein is the reason nitrogenous fertilizers such as urea and ammonium salts are added to farmland. A crop short of nitrogen can still photosynthesise and make carbohydrate, but it cannot build enough protein, so yields fall. The green-chemistry concern is the other half of the story: fertilizer applied in excess is washed by rain into ponds and rivers, where it feeds a heavy growth of algae that uses up the dissolved oxygen and kills fish. Matching the dose to the crop’s actual need is therefore both an agricultural and an environmental decision.
Classify each of the following as a carbohydrate, a protein, a lipid or a nucleic acid, and give the reason for your choice in one line each: (i) cotton fibre, (ii) egg white, (iii) groundnut oil, (iv) the material carrying inherited characters in a cell nucleus, (v) the starch in a wheat grain.
14.2 Carbohydrates
The simple chemical reactions of many carbohydrates show what kind of compounds they are. A carbohydrate is either a polyhydroxy aldehyde or a polyhydroxy ketone, or else a compound that gives rise to such units on hydrolysis.
“Polyhydroxy” simply means that the molecule carries several hydroxyl (-OH) groups. So a carbohydrate molecule has many -OH groups together with one carbonyl group, which may be of the aldehyde type or of the ketone type. Hydrolysis means breaking a large molecule into smaller ones by reaction with water.
Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones, or compounds that give such units on hydrolysis.
Some carbohydrates, such as glucose and fructose, are sweet in taste and are called sugars. The most commonly used sugar is sucrose, which is obtained from sugarcane or sugar beet. The sugar present in milk is called lactose.
Observe the following structural formulae carefully and answer the questions.
\[\mathrm{CHO-(CHOH)_4-CH_2OH}\ \ \text{(glucose)}\] \[\mathrm{CH_2OH-CO-(CHOH)_3-CH_2OH}\ \ \text{(fructose)}\] \[\mathrm{CHO-(CHOH)_3-CH_2OH}\ \ \text{(ribose)}\]
- How many OH groups are present in glucose, fructose and ribose respectively?
- Which other functional groups are present in these three compounds?
Where the Names Come From
The Greek word for sugar is sakkharon. Because of this, carbohydrates are also called saccharides.
The name carbohydrate itself has a different origin. Early chemists noticed that the molecular formulae of many of these compounds could be written as \(\mathrm{C}_x(\mathrm{H_2O})_y\) — that is, as hydrates of carbon. For example, glucose is \(\mathrm{C_6H_{12}O_6}\) or \(\mathrm{C_6(H_2O)_6}\); sucrose is \(\mathrm{C_{12}H_{22}O_{11}}\) or \(\mathrm{C_{12}(H_2O)_{11}}\); and starch is \((\mathrm{C_6H_{10}O_5})_n\) or \([\mathrm{C_6(H_2O)_5}]_n\).
This way of writing the formula is only a bookkeeping device. Water molecules are not actually present inside a glucose molecule; the hydrogen and oxygen atoms simply happen to occur in the 2 : 1 ratio found in water. The name has survived because it is old and convenient, not because it describes the structure.
The word glucose comes from the Greek root glykys, meaning sweet — the same root that gives us “glycerol” and the prefix “glyco-” used all through carbohydrate chemistry. Two ring names in this chapter are borrowed in the same way: a six-membered sugar ring is called a pyranose after the heterocyclic compound pyran, and a five-membered one is called a furanose after furan.
14.2.1 Classification of Carbohydrates
Carbohydrates are classified into three broad groups according to their behaviour on hydrolysis. These are monosaccharides, oligosaccharides and polysaccharides. The complete scheme, with examples, is shown in Fig. 14.1.
Monosaccharides do not hydrolyse further into smaller units of polyhydroxy aldehydes or ketones. They are the building blocks from which all other carbohydrates are made. About twenty different monosaccharides are found in carbohydrates.
A monosaccharide is a carbohydrate that cannot be hydrolysed further into smaller units of polyhydroxy aldehydes or ketones.
Oligosaccharides on hydrolysis yield two to ten units of monosaccharides, and accordingly they are further classified as disaccharides, trisaccharides and so on. Disaccharides are the most common oligosaccharides, and the two monosaccharide units in a disaccharide may be the same or different.
An oligosaccharide is a carbohydrate that yields two to ten monosaccharide units on hydrolysis.
Polysaccharides give a very large number of monosaccharide units on complete hydrolysis. Starch is a common ingredient of food grains, cellulose is a constituent of the cell wall of plant cells, and animals store polysaccharides in their body in the form of glycogen.
A polysaccharide is a carbohydrate that yields a very large number of monosaccharide units on complete hydrolysis.
Monosaccharides vs Oligosaccharides vs Polysaccharides
| Behaviour on hydrolysis |
Do not hydrolyse further |
Yield 2 to 10 monosaccharide units |
Yield a very large number of units |
| Sub-classes |
— |
Disaccharides, trisaccharides, tetrasaccharides |
— |
| Examples |
Glucose, fructose, ribose |
Sucrose, maltose, lactose, raffinose, stachyose |
Starch, glycogen, cellulose |
| Taste |
Sweet |
Usually sweet |
Not sweet |
On complete hydrolysis, a carbohydrate X gives one molecule of glucose, one of fructose and one of galactose. Another carbohydrate Y gives several hundred molecules of glucose only. Classify X and Y, and state to which sub-class each belongs.
14.2.2 Nomenclature of Monosaccharides
According to the IUPAC system of nomenclature, the general name for a monosaccharide is glycose. The particular name given to a monosaccharide tells us two things: the kind of carbonyl group it carries, and the number of carbon atoms in it.
A monosaccharide with one aldehydic carbonyl group is called an aldose, while one with a ketonic carbonyl group is called a ketose.
An aldose is a monosaccharide containing one aldehydic carbonyl group; a ketose is a monosaccharide containing one ketonic carbonyl group.
These names are further modified in accordance with the total number of carbon atoms in the monosaccharide. A monosaccharide with three carbons is a triose, with four a tetrose, with five a pentose and with six a hexose; the two parts of the name are then joined together.
For example, glucose (\(\mathrm{C_6H_{12}O_6}\)) is an aldose with six carbons, and is thereby an aldohexose. Fructose (\(\mathrm{C_6H_{12}O_6}\)) is a ketose with six carbons, and is thereby a ketohexose.
Give IUPAC names to the following monosaccharides.
- \(\mathrm{CHO-CHOH-CH_2OH}\)
- \(\mathrm{CHO-(CHOH)_3-CH_2OH}\)
- \(\mathrm{CH_2OH-CO-(CHOH)_3-CH_2OH}\)
14.2.3 Glucose
Glucose occurs in nature in the free as well as in the combined state. It can be obtained from sucrose or from starch by acid catalysed hydrolysis, as shown below.
Preparation of Glucose from Sucrose
Sucrose is hydrolysed by warming it with dilute hydrochloric acid or sulfuric acid for about two hours. This hydrolysis converts sucrose into a mixture of glucose and fructose.
\[\mathrm{C_{12}H_{22}O_{11}} + \mathrm{H_2O} \xrightarrow{\ \mathrm{H^{\oplus}},\ \Delta\ } \mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6}\]
Here \(\mathrm{C_{12}H_{22}O_{11}}\) is sucrose, and the two products are glucose and fructose. The symbol \(\mathrm{H^{\oplus}}\) over the arrow shows that the reaction is catalysed by acid, and \(\Delta\) shows that the mixture is warmed.
Glucose is then separated from fructose by adding ethanol during cooling. Glucose, being almost insoluble in alcohol, crystallizes out first. The solution is filtered to obtain crystals of glucose. The whole sequence is summarised in FC 14.2.
Preparation of Glucose from Starch
Commercially, glucose is obtained by hydrolysis of starch, by boiling it with dilute sulfuric acid at 393 K under 2 to 3 atm pressure.
\[(\mathrm{C_6H_{10}O_5})_n + n\,\mathrm{H_2O} \xrightarrow{\ \mathrm{H^{\oplus}},\ 393\ \text{K},\ 2\text{–}3\ \text{atm}\ } n\,\mathrm{C_6H_{12}O_6}\]
Starch is a polymer, so one starch molecule containing \(n\) glucose units needs \(n\) molecules of water and releases \(n\) molecules of glucose. The higher temperature and pressure are needed because the linkages in a polysaccharide are harder to break than the single linkage in a disaccharide.
Almost all the glucose used in confectionery, in soft drinks and in glucose powders sold as an energy source is made industrially by this starch route, using maize or other cheap starch-rich grain. A handling point matters here: whenever dilute sulfuric acid has to be prepared for such a hydrolysis, the concentrated acid is added slowly to water with stirring, never water to the acid. Mixing releases a great deal of heat, and adding water to concentrated acid can make the small amount of liquid boil and spit acid out of the vessel.
Calculate the mass of glucose obtained by the complete hydrolysis of 324 g of starch. (Molar mass of one \(\mathrm{C_6H_{10}O_5}\) unit = 162 g mol⁻¹; molar mass of glucose = 180 g mol⁻¹.)
From the equation above, each \(\mathrm{C_6H_{10}O_5}\) unit of starch takes up one molecule of water and gives one molecule of glucose. So the calculation can be done per unit.
Number of moles of \(\mathrm{C_6H_{10}O_5}\) units
\[n = \frac{324\ \text{g}}{162\ \text{g mol}^{-1}} = 2\ \text{mol}\]
Since 1 mol of units gives 1 mol of glucose, 2 mol of units give 2 mol of glucose.
\[\text{Mass of glucose} = 2\ \text{mol} \times 180\ \text{g mol}^{-1} = 360\ \text{g}\]
Notice that the product weighs more than the starting material. That is not an error: the extra 36 g is the mass of the 2 mol of water taken up during hydrolysis.
171 g of sucrose is completely hydrolysed by warming with dilute hydrochloric acid. Calculate the mass of glucose and the mass of fructose formed. (Molar mass of sucrose = 342 g mol⁻¹, of glucose and of fructose = 180 g mol⁻¹ each.)
14.2.4 Structure and Properties of Glucose
Glucose has an aldohexose structure. In other words, a glucose molecule contains one aldehydic — that is, formyl — group, and the remaining five carbons carry one hydroxyl (-OH) group each. The six carbons in glucose form one straight chain.
This aldohexose structure of glucose was not assumed; it was established on the basis of the following chemical properties. Each observation rules out certain possible structures and supports one particular arrangement, as summarised in FC 14.3.
3. One Carbonyl Group Is Present
A glucose molecule contains one carbonyl group. This was inferred from the observation that glucose forms an oxime by reaction with hydroxylamine, and gives a cyanohydrin on reaction with hydrogen cyanide.
\[\mathrm{CHO-(CHOH)_4-CH_2OH} \xrightarrow{\ \mathrm{NH_2OH},\ -\mathrm{H_2O}\ } \mathrm{CH{=}N{-}OH\ \text{-}(CHOH)_4-CH_2OH}\]
\[\mathrm{CHO-(CHOH)_4-CH_2OH} \xrightarrow{\ \mathrm{HCN}\ } \mathrm{CH(OH)(CN)-(CHOH)_4-CH_2OH}\]
Both of these are reactions that only a carbonyl group gives, and only one molecule of each reagent is taken up. That fixes the number of carbonyl groups at one.
4. The Carbonyl Group Is an Aldehyde
This was inferred from the observation that glucose gets oxidised to a six-carbon monocarboxylic acid called gluconic acid on reaction with bromine water, which is a mild oxidizing agent.
\[\mathrm{CHO-(CHOH)_4-CH_2OH} \xrightarrow{\ (\mathrm{O}),\ \mathrm{Br_2}\ \text{water}\ } \mathrm{COOH-(CHOH)_4-CH_2OH}\]
Only an aldehyde group is oxidised so easily; a ketone group is not. The product still has six carbons and only one -COOH group, which shows that the carbonyl group was at the end of the chain.
bromine water and dilute nitric acid (Bromine water, \(\mathrm{Br_2}\) in water, is a mild oxidising agent → it attacks only the aldehyde group of a sugar and converts -CHO to -COOH, leaving the alcoholic -OH groups alone → used to show that a sugar’s carbonyl group is an aldehyde, and to prepare gluconic acid. Dilute nitric acid is a stronger oxidising agent → it oxidises both the terminal -CHO and the terminal primary -CH₂OH to -COOH → gives a dicarboxylic acid, and so is used to prove that a primary alcoholic group is present at the other end of the chain.)
An alcoholic compound was found to have molecular mass of 90 u. It was acetylated. The molecular mass of the acetyl derivative was found to be 174 u. How many alcoholic (-OH) groups must be present in the original compound?
In an acetylation reaction, the H atom of an (-OH) group is replaced by an acetyl group (\(\mathrm{-COCH_3}\)). This results in an increase in molecular mass by \([(12+16+12+3\times1)-1]\), that is, 42 u.
In the given alcohol,
\[\text{increase in molecular mass} = 174\ \text{u} - 90\ \text{u} = 84\ \text{u}\]
\[\therefore\ \text{Number of -OH groups} = \frac{84\ \text{u}}{42\ \text{u}} = 2\]
Each -OH group that reacts adds a fixed 42 u to the molecule, so the total gain in mass divided by 42 u directly counts the -OH groups. This is exactly the reasoning used for glucose in point 5 below.
What are the products of reaction of
- \(\mathrm{CH_3-CO-CH_3}\) with \(\mathrm{NH_2-OH}\)?
- \(\mathrm{CH_3-CHO}\) with HCN?
- \(\mathrm{CH_3-OH}\) with \(\mathrm{CH_3-CO-O-CO-CH_3}\)?
5. Glucose Contains Five Hydroxyl Groups
This was inferred from the observation that glucose reacts with five moles of acetic anhydride to form glucose pentaacetate.
\[\mathrm{CHO-(CHOH)_4-CH_2OH} + 5\,(\mathrm{CH_3CO})_2\mathrm{O} \rightarrow \mathrm{CHO-(CH-O-CO-CH_3)_4-CH_2-O-CO-CH_3} + 5\,\mathrm{CH_3COOH}\]
The products are glucose pentaacetate and acetic acid. As glucose is a stable compound, it was further inferred that the five hydroxyl groups are bonded to five different carbon atoms in the glucose molecule.
6. Glucose Contains One Primary Alcoholic Group
Glucose contains one primary alcoholic (\(\mathrm{-CH_2OH}\)) group. This was inferred from the following observation: glucose and gluconic acid both, on oxidation with dilute nitric acid, give the same dicarboxylic acid, called saccharic acid.
\[\mathrm{CHO-(CHOH)_4-CH_2OH} \xrightarrow{\ (\mathrm{O}),\ \mathrm{HNO_3}\ } \mathrm{COOH-(CHOH)_4-COOH}\]
\[\mathrm{COOH-(CHOH)_4-CH_2OH} \xrightarrow{\ (\mathrm{O}),\ \mathrm{HNO_3}\ } \mathrm{COOH-(CHOH)_4-COOH}\]
If both compounds end up as the same dicarboxylic acid, the group at the far end of the chain must be one that can itself be oxidised to -COOH. A primary alcoholic \(\mathrm{-CH_2OH}\) group does exactly that, and this fixes the last carbon of the chain.
A monosaccharide gives an oxime with hydroxylamine but is not** oxidised by bromine water. What does each of these two observations tell you about its carbonyl group, and to which class of monosaccharide does the compound belong?**
14.2.5 Optical Isomerism in Glucose
The structural formula of glucose shows that it contains four chiral carbon atoms. You have learnt that every chiral carbon can have two distinct spatial arrangements of groups around it (section 10.5.1). In other words, two distinct configurations are possible for each of the four chiral carbons of glucose.
A chiral carbon is one attached to four different groups, so that its mirror image cannot be superimposed on it. The stereostructure of glucose is therefore only one out of several possible stereostructures of an aldohexose.
A structural formula containing ‘n’ number of chiral carbons can have a maximum of ‘\(2^n\)’ numbers of stereostructures or optical isomers. An aldohexose, therefore, can exist as sixteen (\(2^4 = 16\)) optical isomers, and glucose is one of them.
A molecule containing n chiral carbon atoms can exist in a maximum of \(2^n\) stereoisomers (optical isomers).
- Write the structural formula of glucose showing all the bonds in the molecule.
- Number all the carbons in the molecule, giving number 1 to the (-CHO) carbon.
- Mark the chiral carbons in the molecule with an asterisk (*).
- How many chiral carbons are present in glucose?
- What are the ways to represent the three dimensional structure of an organic molecule?
- How is a Fischer projection formula drawn?
On the basis of very elaborate chemical evidence and measurement of the optical activity of the various chemicals involved, Emil Fischer, a German Nobel laureate (1902), determined the configuration of the four chiral carbons (C-2, C-3, C-4, C-5) in glucose. The Fischer projection formulae of glucose (I), gluconic acid (II) and saccharic acid (III) are shown in Fig. 14.2.
Rotation of Plane-Polarised Light
Glucose is an optically active compound and has its specific rotation, \([\alpha]_D^{20}\), equal to \(+52.7^\circ\). Due to its dextrorotation, glucose is also called dextrose. The designations (+)-glucose or d-glucose imply the dextrorotatory nature of glucose.
D-glucose is another designation of glucose, which is more common. This designation indicates the configuration of glucose rather than the sign of its optical rotation. The capital letter D and the sign (+) are therefore two separate pieces of information, and it is a common examination trap to treat them as the same thing.
specific rotation \([\alpha]_D^{20}\) (Measured in degrees (°); the subscript D means the sodium D line is used as the light source and the superscript 20 means the measurement is made at 20 °C. Sign convention: a positive value means the substance rotates the plane of polarised light clockwise, that is, it is dextrorotatory; a negative value means anticlockwise rotation, that is, laevorotatory. The sign must be determined by experiment; it cannot be predicted from the structure.)
The D/L Configuration System
The prefix D- or L- in the name of a compound indicates the relative configuration of a stereoisomer. It refers to a particular enantiomer of glyceraldehyde. Glyceraldehyde has one chiral carbon (C-2) and exists as two enantiomers. These are represented by two Fischer projection formulae, shown in Fig. 14.3.
Conventionally, (+)-glyceraldehyde is represented by the Fischer projection formula having the OH group attached to C-2 on the right side (IV), and this configuration is denoted by the symbol ‘D’. Similarly, the configuration of (−)-glyceraldehyde (V) is denoted by the symbol ‘L’.
All the compounds which can be correlated by a series of chemical reactions to (+)-glyceraldehyde are said to have D-configuration. The compounds which are chemically correlated to (−)-glyceraldehyde are said to have L-configuration. This is the system of relative configuration of chiral compounds.
In the D/L system, a compound is assigned the D-configuration if it can be correlated by a series of chemical reactions to (+)-glyceraldehyde, and the L-configuration if it can be correlated to (−)-glyceraldehyde. The system gives relative configuration only, and says nothing about the direction in which the compound rotates plane-polarised light.
A monosaccharide is assigned D/L configuration on the basis of the configuration of the lowest chiral carbon in its Fischer projection formula. “Lowest” here means the chiral carbon nearest the bottom of the Fischer projection, that is, the one farthest from the carbonyl group. The D-configuration of (+)-glucose is illustrated in Fig. 14.4.
The D/L system had to be built on a relative comparison for a very practical reason: optical rotation is an experimentally measurable property of a compound, but the actual arrangement of groups around a chiral carbon is difficult to observe by any simple experiment. That gap was closed only in 1951, when X-ray crystallographic studies of (+)-sodium rubidium tartarate established its configuration. This was the first instance of determining absolute configuration.
Assign D/L configuration to the following monosaccharides.
i. Threose: Fischer projection with C-1 CHO at the top, C-2 with HO on the left, C-3 with H on the left and OH on the right, and C-4 \(\mathrm{CH_2OH}\) at the bottom.
ii. Ribose: Fischer projection with C-1 CHO at the top, C-2, C-3 and C-4 each with HO on the left, and C-5 \(\mathrm{CH_2OH}\) at the bottom.
D/L configuration is assigned to the Fischer projection formula of a monosaccharide on the basis of the lowest chiral carbon.
- Threose has two chiral carbons, C-2 and C-3. The given Fischer projection formula of threose has the -OH group at the lowest C-3 chiral carbon on the right side.
\[\therefore\ \text{It is D-threose.}\]
- Ribose has three chiral carbons, C-2, C-3 and C-4. The given Fischer projection formula of ribose has the -OH group at the lowest C-4 chiral carbon on the left side.
\[\therefore\ \text{It is L-ribose.}\]
A ketohexose is drawn as a Fischer projection with \(\mathrm{CH_2OH}\) at the top, a C=O at C-2, and -OH groups on the left at C-3, on the right at C-4 and on the right at C-5. (i) How many chiral carbons does it contain? (ii) What is the maximum number of optical isomers possible for this molecule? (iii) Assign it the D or the L configuration, giving your reason.
14.2.6 Ring Structure of Glucose
On the basis of chemical evidence, the stereostructure of D-glucose was represented by the Fischer projection formula I (Fig. 14.2 and Fig. 14.4). Glucose, however, was found to exhibit some more chemical properties which could not be explained on the basis of the structure I. It was necessary to write another structure for glucose which will explain all the properties. The ring structure of glucose fulfils this requirement.
Glucose is found to have two cyclic structures (VI and VII) which are in equilibrium with each other through the open chain structure (I) in aqueous solution, as shown in Fig. 14.5.
14.2.7 Reducing Nature of Glucose
The hemiacetal group of the glucopyranose structure is a potential aldehyde group. Although the -CHO group is not visible in the ring formula, the ring can open at any moment to release it, because all three forms are in equilibrium in solution.
This potential aldehyde group imparts reducing properties to glucose. Thus, glucose gives a positive Tollens test and a positive Fehling test (Section 12.8.1 a).
A reducing sugar is a sugar that contains a free aldehyde or potential aldehyde (hemiacetal) group and therefore reduces Tollens’ reagent and Fehling’s solution.
Tollens’ reagent and Fehling’s solution (Tollens’ reagent is ammoniacal silver nitrate → it needs an aldehyde or a potential aldehyde group in the substance being tested → a positive result appears as a bright silver mirror on the inner wall of the test tube. Fehling’s solution contains copper(II) ions in alkaline tartrate solution → it is used the same way, on warming → a positive result appears as a red-brown precipitate of copper(I) oxide. Both are used in this chapter for one purpose: to decide whether a given sugar is reducing or non-reducing.)
14.2.8 Representation of Fructose Structure
Fructose (\(\mathrm{C_6H_{12}O_6}\)) is a laevorotatory ketohexose. Fructose is also called laevulose due to its laevorotation, \([\alpha]_D^{20} = -92.4^\circ\).
Being an α-hydroxy keto compound, fructose is a reducing sugar. Its carbonyl group is a ketone, yet the -OH group on the neighbouring carbon allows the molecule to behave as a reducing agent in the alkaline conditions of the Tollens and Fehling tests.
In the free state, fructose exists as a mixture of fructopyranose (major) and fructofuranose. In the combined state, fructose is found in the form of the fructofuranose ring structure (as in sucrose, see section 14.2.9).
The name furanose is given by analogy with furan, a five-membered heterocyclic compound. Figure 14.7 shows representations of the open chain structure of fructose and the ring structures of α- and β-anomers of fructofuranose. The ring structure of fructose is a hemiketal (section 12.8.2 c) — the counterpart of a hemiacetal, formed when the alcohol adds to a ketone carbonyl instead of an aldehyde carbonyl.
Q4. Glucose is written as D-(+)-glucose and fructose as D-(-)-fructose. Both carry the letter D, but one sign is positive and the other negative. What does the letter tell us, and what does the sign tell us — and why are they not the same information?
A. The capital letter D describes configuration. It says that the lowest chiral carbon in the Fischer projection has its -OH on the right, matching (+)-glyceraldehyde. Both glucose and fructose satisfy this, so both are D-sugars. The sign in brackets describes something entirely different — the direction in which the compound actually rotates plane-polarised light, which is found only by experiment with a polarimeter. Glucose rotates it clockwise (\([\alpha]_D^{20} = +52.7^\circ\)) and fructose anticlockwise (\([\alpha]_D^{20} = -92.4^\circ\)). Configuration and rotation are not linked by any simple rule, which is exactly why two separate symbols are needed.
Fructose has no aldehyde group at all, yet it gives a positive Fehling test. Explain this observation, and state which ring size fructose adopts when it is present in sucrose.
14.2.9 Disaccharides
Disaccharides give rise to two units of the same or different monosaccharides on hydrolysis with dilute acids or specific enzymes. The two monosaccharide units are linked together by an ether oxide linkage (-O-), which is termed a glycosidic linkage in carbohydrate chemistry.
A glycosidic linkage is formed by the removal of a water molecule by reaction of two hydroxyl (-OH) groups from two monosaccharide units. At least one of the two monosaccharide units must use its anomeric hydroxyl group in the formation of the glycosidic linkage.
A glycosidic linkage is the ether (-O-) linkage joining two monosaccharide units, formed by the loss of one water molecule between two of their hydroxyl groups, at least one of which must be the anomeric hydroxyl group.
The three most common disaccharides are sucrose, maltose and lactose.
a. Sucrose
Sucrose (\(\mathrm{C_{12}H_{22}O_{11}}\)) is dextrorotatory (\(+66.5^\circ\)). On hydrolysis with dilute acid or an enzyme called invertase, sucrose gives an equimolar mixture of D-(+) glucose and D-(-) fructose.
\[\mathrm{C_{12}H_{22}O_{11}} + \mathrm{H_2O} \xrightarrow{\ \mathrm{H^{\oplus}}\ \text{or invertase}\ } \mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6}\]
Since the laevorotation of fructose (\(-92.4^\circ\)) is larger than the dextrorotation of glucose (\(+52.7^\circ\)), the hydrolysis product has net laevorotation. The rotation of the solution therefore changes sign as the reaction proceeds — from \(+66.5^\circ\) for the original sucrose to a negative value for the product mixture.
Hence hydrolysis of sucrose is also called inversion of sucrose, and the product is called invert sugar.
The structure of sucrose contains a glycosidic linkage between C-1 of α-glucose and C-2 of β-fructose (Fig. 14.8).
Since the potential aldehyde and ketone groups of both the monosaccharide units are involved in the formation of the glycosidic bond, sucrose is a non reducing sugar. Neither ring can open to release a carbonyl group, so there is nothing left in the molecule for Tollens’ reagent or Fehling’s solution to react with.
Make models corresponding to the two Haworth formulae of sucrose in Fig. 14.8. Check that both are identical.
Invert sugar is commercially available as invert syrup. It is used as a sweetener in bakery and confectionery products and also in fruit preserves and beverages. It is sweeter than sucrose and glucose. It is resistant to crystallization and promotes retention of moisture, enhances flavour and texture, and also prolongs shelf life.
68.4 g of sucrose is completely hydrolysed with dilute acid. Calculate the mass of glucose and the mass of fructose obtained. (Molar mass of sucrose = 342 g mol⁻¹; molar mass of glucose = molar mass of fructose = 180 g mol⁻¹.)
From the equation above, 1 mol of sucrose gives 1 mol of glucose and 1 mol of fructose.
Number of moles of sucrose
\[n = \frac{68.4\ \text{g}}{342\ \text{g mol}^{-1}} = 0.2\ \text{mol}\]
Therefore 0.2 mol of glucose and 0.2 mol of fructose are formed.
\[\text{Mass of glucose} = 0.2\ \text{mol} \times 180\ \text{g mol}^{-1} = 36\ \text{g}\]
\[\text{Mass of fructose} = 0.2\ \text{mol} \times 180\ \text{g mol}^{-1} = 36\ \text{g}\]
The word “equimolar” in the description of the products is the key to this calculation: the two sugars are formed in a 1 : 1 mole ratio, and because they happen to have the same molar mass, their masses come out equal as well.
b. Maltose
Maltose (\(\mathrm{C_{12}H_{22}O_{11}}\)) is a disaccharide made of two units of D-glucose. The glycosidic bond in maltose is formed between C-1 of one glucose ring and C-4 of the other.
The glucose ring which uses its hydroxyl group at C-1 is α-glucopyranose. Hence the linkage is called an α-1,4-glycosidic linkage.
The hemiacetal group at C-1 of the second ring is not involved in the glycosidic linkage. Hence maltose is a reducing sugar — that free hemiacetal can still open into an aldehyde group. Maltose gives glucose on hydrolysis with dilute acids or the enzyme maltase.
\[\mathrm{C_{12}H_{22}O_{11}} + \mathrm{H_2O} \xrightarrow{\ \mathrm{H^{\oplus}}\ \text{or maltase}\ } 2\,\mathrm{C_6H_{12}O_6}\]
Figure 14.9 shows the Haworth formula of maltose.
c. Lactose
Lactose (\(\mathrm{C_{12}H_{22}O_{11}}\)) is a disaccharide present in milk. It is formed from two monosaccharide units, namely D-galactose and D-glucose.
The glycosidic linkage is formed between C-1 of β-D-galactose and C-4 of glucose. Therefore the linkage in lactose is called a β-1,4-glycosidic linkage.
The hemiacetal group at C-1 of the glucose unit is not involved in the glycosidic linkage but is free. Hence lactose is a reducing sugar. Figure 14.10 shows the Haworth formula of lactose.
Sucrose vs Maltose vs Lactose
| Molecular formula |
\(\mathrm{C_{12}H_{22}O_{11}}\) |
\(\mathrm{C_{12}H_{22}O_{11}}\) |
\(\mathrm{C_{12}H_{22}O_{11}}\) |
| Monosaccharide units |
α-glucose + β-fructose |
Two units of D-glucose |
D-galactose + D-glucose |
| Glycosidic linkage |
α, β-1,2 (C-1 of glucose to C-2 of fructose) |
α-1,4 |
β-1,4 (C-1 of galactose to C-4 of glucose) |
| Free hemiacetal group left? |
No — both anomeric groups are used |
Yes, at C-1 of the second glucose ring |
Yes, at C-1 of the glucose unit |
| Reducing or non-reducing |
Non reducing |
Reducing |
Reducing |
| Hydrolysing enzyme |
Invertase |
Maltase |
— |
| Chief source |
Sugarcane, sugar beet |
— |
Milk |
Q1. Glucose, maltose and sucrose are all sugars, and sucrose is even built out of glucose. Why then do glucose and maltose give a positive Tollens test, while sucrose gives a negative one?
A. The test does not ask what a sugar is made of; it asks whether the molecule still has a free aldehyde or potential aldehyde group. In glucose, the ring is a hemiacetal that can open to give a -CHO group. In maltose, only one of the two glucose units uses its anomeric -OH in the glycosidic bond, so the second ring still has a free hemiacetal at C-1 and can open in the same way. In sucrose, both units use their anomeric groups in the C-1 to C-2 linkage. Neither ring can open, no carbonyl group is available, and the Tollens test comes out negative.
- Is galactose an aldohexose or a ketohexose?
- Which carbon in galactose has a different configuration compared to glucose?
- Draw Haworth formulae of α-D-galactose and β-D-galactose.
- Which disaccharides among sucrose, maltose and lactose is/are expected to give a positive Fehling test?
- What are the expected products of hydrolysis of lactose?
14.2.10 Polysaccharides
Polysaccharides are formed by linking a large number of monosaccharide units by glycosidic linkages. The same kind of linkage you met in disaccharides is simply repeated hundreds or thousands of times.
Starch, cellulose and glycogen are the most common natural polysaccharides.
Starch is the storage carbohydrate of plants and an important nutrient for humans and other animals. Cellulose is the main constituent of the cell wall of plant and bacterial cells; it is also the main constituent of wood and cotton. Glycogen constitutes the storage carbohydrate of animals and is present in liver, muscles and brain; it is also found in yeast and fungi.
a. Starch
Starch is a polymer of α-D-glucose. Starch has two components, namely amylose (15-20%) and amylopectin (80-85%).
Amylose is soluble in water and forms a blue coloured complex with iodine. It contains 200-1000 α-glucose units linked by α-1,4-glycosidic linkages, giving rise to an unbranched chain of variable length (Fig. 14.11).
Amylopectin is the water insoluble component of starch, which forms a blue-violet coloured complex with iodine. It is a branched chain polysaccharide. In amylopectin, chains are formed by α-1,4-glycosidic linkages between α-glucose units, whereas branches are formed by α-1,6-glycosidic linkages (Fig. 14.12).
When you chew plain bread, chapati or bhaakari for a long time, it tastes sweet. What could be the reason?
b. Cellulose
Cellulose is a straight chain polysaccharide of β-glucose units linked by β-1,4-glycosidic bonds. Note the single but decisive difference from starch: the glucose units are the β form, not the α form, and everything else follows from that.
Chemical hydrolysis of cellulose requires the use of concentrated strong acids at high temperature and pressure. This implies that the β-1,4-glycosidic bond is very strong and difficult to hydrolyse.
Humans do not have enzymes which can hydrolyse this linkage. Hence cellulose cannot be digested by human beings; it serves as the fibrous content of food, useful for bowel movement. Figure 14.13 shows the Haworth formula of cellulose.
The β-1,4 linkage is not unbreakable — it is only unbreakable by us. The symbiotic bacteria in the guts of insects called termites have enzymes that can hydrolyse the β-1,4-glycosidic linkage in cellulose, which is how a termite lives on wood. Grazing animals such as cattle manage the same trick with microorganisms in their stomachs. For human beings, the undigested cellulose in vegetables and whole grains is what nutritionists call dietary fibre or roughage, and its value lies precisely in the fact that we cannot break it down.
c. Glycogen
Glycogen has its structure similar to that of amylopectin, but it is more highly branched. The heavy branching suits its function: many chain ends are exposed at once, so glucose can be released quickly when the animal needs energy.
Starch vs Cellulose vs Glycogen
| Monomer unit |
α-D-glucose |
β-glucose |
α-D-glucose |
| Linkage |
α-1,4 (amylose); α-1,4 with α-1,6 branches (amylopectin) |
β-1,4 |
As in amylopectin, but more highly branched |
| Chain shape |
Amylose unbranched; amylopectin branched |
Straight chain |
Highly branched |
| Occurrence |
Storage carbohydrate of plants; food grains |
Cell wall of plant and bacterial cells; wood, cotton |
Storage carbohydrate of animals; liver, muscles, brain; also yeast and fungi |
| Digestible by humans |
Yes |
No |
Yes |
Q2. Starch and cellulose are both long chains made only of glucose units. Why can we digest starch easily but not cellulose at all, even though the building block is the same?
A. The building block is not quite the same, and the linkage is not the same at all. Starch is built from α-D-glucose units joined by α-1,4 linkages, while cellulose is built from β-glucose units joined by β-1,4 linkages. The β-1,4 bond is very strong — its chemical hydrolysis needs concentrated strong acid at high temperature and pressure — and the human body has no enzyme shaped to break it. Enzymes are highly specific, so an enzyme that splits the α linkage in starch simply does not fit the β linkage in cellulose. The cellulose therefore passes through us undigested as dietary fibre.
A sample of starch is treated with iodine solution and gives two colours from its two components. (i) Name the two components and state the colour each gives with iodine. (ii) State which of the two is soluble in water. (iii) Write the type of glycosidic linkage found at a branch point, and name the polysaccharide of animals that has the same branched pattern but to a greater degree.
14.3 Proteins
- What is the product of reaction of acetic acid with ammonia?
- Write the structural formula of N-methyl acetamide. What is the name of the functional group in this compound?
- What are the nitrogenous nutrients in human diet?
Proteins are the fundamental structural materials of animal bodies. Proteins in the form of enzymes play a prime role in all the physiological reactions taking place inside an organism.
Nutritional sources of proteins are milk, pulses, nuts, fish, meat, and similar foods. Chemically, proteins are polyamides, which are high molecular weight polymers of the monomer units called α-amino acids.
An amide is the linkage \(\mathrm{-CO-NH-}\) formed between a carboxyl group and an amino group. A polyamide is simply a long chain in which this linkage is repeated many times over — which is exactly what a protein molecule is.
The name protein is derived from the Greek word proteios, which means “primary” or “of prime importance”. The naming was apt: no other class of biomolecule does so many different jobs in the body.
14.3.1 α-Amino Acids
Proteins on complete hydrolysis give rise to a mixture of α-amino acids. An α-amino acid is a carboxylic acid having an amino (\(\mathrm{-NH_2}\)) group bonded to the α-carbon, that is, the carbon next to the carboxyl (-COOH) group (Fig. 14.14).
\[\mathrm{R-CH(NH_2)-COOH}\]
An α-amino acid is a carboxylic acid in which an amino (\(\mathrm{-NH_2}\)) group is attached to the α-carbon — the carbon atom directly next to the carboxyl (-COOH) group.
The α-carbon in all the α-amino acids (except glycine) is chiral. In glycine the side chain is only a hydrogen atom, so the α-carbon carries two identical hydrogens and cannot be chiral.
It is found that the α-carbon in α-amino acids obtained by hydrolysis of proteins has the ‘L’ configuration. The L-α-amino acids are represented by the Fischer projection formula as shown in Fig. 14.14. Recall from section 14.2.5 that L and D describe configuration, not the direction of optical rotation.
Side Chains Decide the Type
The symbol ‘R’ in the structure of α-amino acids represents the side chain and may contain additional functional groups. It is this side chain, and nothing else, that makes one amino acid different from another.
If ‘R’ contains a carboxyl (-COOH) group, the amino acid is an acidic amino acid. If ‘R’ contains an amino group (1°, 2° or 3°), it is called a basic amino acid. The other amino acids, having a neutral or no functional group in ‘R’, are called neutral amino acids.
α-Amino acids have trivial names and are generally represented by three letter symbols, or sometimes by a one letter symbol. Table 14.1 lists the twenty α-amino acids, often referred to simply as amino acids, commonly found in proteins, with their symbols and also their types as neutral, acidic or basic.
Ten α-amino acids from this list cannot be synthesised in the human body and have to be obtained through diet. These are called essential amino acids and are marked with an asterisk (*) in Table 14.1.
Essential amino acids are those α-amino acids which the human body cannot synthesise and which must therefore be supplied through the diet.
Table 14.1: Natural α-amino acids : L-RCH(NH₂)COOH (* Essential α-amino acids)
| Neutral α-amino acids |
1. Glycine |
H- |
Gly |
G |
|
2. Alanine |
CH₃- |
Ala |
A |
|
3. Valine* |
Me₂CH- |
Val |
V |
|
4. Leucine* |
Me₂CH-CH₂- |
Leu |
L |
|
5. Isoleucine* |
CH₃-CH₂-CH(Me)- |
Ile |
I |
|
6. Asparagine |
H₂N-CO-CH₂- |
Asn |
N |
|
7. Glutamine |
H₂N-CO-CH₂-CH₂- |
Gln |
Q |
|
8. Serine |
HO-CH₂- |
Ser |
S |
|
9. Threonine* |
CH₃-CHOH- |
Thr |
T |
|
10. Cysteine |
HS-CH₂- |
Cys |
C |
|
11. Methionine* |
Me-S-CH₂-CH₂ |
Met |
M |
|
12. Phenylalanine* |
Ph-CH₂- |
Phe |
F |
|
13. Tyrosine |
p-HO-C₆H₄-CH₂- |
Tyr |
Y |
|
14. Tryptophan* |
-CH₂- attached at position 3 of indole (benzene ring fused to a pyrrole ring carrying NH) |
Trp |
W |
|
15. Proline |
entire structure: a five-membered ring containing NH, with -COOH on the carbon next to NH |
Pro |
P |
| Acidic |
16. Aspartic acid |
HOOC-CH₂- |
Asp |
D |
|
17. Glutamic acid |
HOOC-CH₂-CH₂- |
Glu |
G |
| Basic |
18. Lysine* |
H₂N-(CH₂)₄- |
Lys |
K |
|
19. Arginine* |
HN = C(NH₂) - NH - (CH₂)- |
Arg |
R |
|
20. Histidine* |
-CH₂- attached to the imidazole ring (five-membered ring with two nitrogen atoms, one carrying H) |
His |
H |
Tryptophan and histidine have the structures (I) and (II) respectively (see Table 14.1). Classify them into neutral/acidic/basic α-amino acids and justify your answer. (Hint: Consider involvement of the lone pair in resonance.)
The idea of an essential amino acid explains an old habit of the Indian kitchen. Cereals such as wheat and rice are low in lysine, while pulses are low in methionine; each is short of an essential amino acid the other supplies. Eaten together — dal with rice, rajma with chapati — the two provide the full set between them. This is also why growing children and people recovering from illness are advised to take milk, pulses and eggs rather than cereals alone.
Classify each of the following α-amino acids as acidic, basic or neutral, using the side chain R given in Table 14.1, and state which of them is an essential amino acid: (i) aspartic acid, (ii) lysine, (iii) serine.
The rule to apply is that the type is decided only by the extra functional group present in the side chain R.
Aspartic acid has \(\mathrm{R = HOOC-CH_2-}\). The side chain carries an extra carboxyl group, which can donate a proton. Therefore aspartic acid is an acidic amino acid. It is not marked with an asterisk in Table 14.1, so it is not essential.
Lysine has \(\mathrm{R = H_2N-(CH_2)_4-}\). The side chain carries an extra amino group, whose nitrogen lone pair can accept a proton. Therefore lysine is a basic amino acid. It is marked with an asterisk, so it is an essential amino acid.
Serine has \(\mathrm{R = HO-CH_2-}\). The side chain carries only an alcoholic -OH group, which is neither appreciably acidic nor basic. Therefore serine is a neutral amino acid, and it is not essential.
Compare the molecular masses of the following compounds and explain the observed melting points.
| CH₃-CH(NH₂)-COOH |
89 |
293.5 °C |
| C₅H₁₁-NH₂ |
87 |
−55 °C |
| C₃H₇-COOH |
88 |
−7.9 °C |
The Zwitter Ion Structure
α-Amino acids are high melting, water soluble crystalline solids, unlike simple amines or carboxylic acids of comparable molecular mass. These properties are due to a peculiar structure called the zwitter ion structure of α-amino acids.
An α-amino acid molecule contains both an acidic carboxyl (-COOH) group as well as a basic amino (\(\mathrm{-NH_2}\)) group. Proton transfer from the acidic group to the basic group of the amino acid forms a salt, which is a dipolar ion called a zwitter ion (Fig. 14.15).
\[\mathrm{R-CH(NH_2)-COOH} \rightarrow \mathrm{R-CH(\overset{+}{N}H_3)-COO^{-}}\]
A zwitter ion is the dipolar ion of an amino acid, formed by transfer of the proton of the carboxyl group to the amino group of the same molecule, so that the ion carries a positive and a negative charge but no net charge.
The odd-looking name is borrowed straight from German, where Zwitter means “hybrid” — a fitting word for an ion that is positive at one end and negative at the other at the same time.
This structure explains the properties listed above. Because the zwitter ion is really a salt, the crystal is held together by strong electrostatic forces between full charges, which need a high temperature to break — hence the very high melting point. The charges also attract water molecules strongly, which is why amino acids dissolve readily in water. A simple amine or carboxylic acid of similar molecular mass has no such charges and therefore melts far below room temperature.
Effect of pH
An amino acid can exist in different forms depending upon the pH of the aqueous solution in which it is dissolved. Consider, for example, the zwitter ion and the other forms of alanine (Fig. 14.16).
In strongly acidic solution (pH ≤ 2) there are plenty of protons available, so the \(\mathrm{-COO^-}\) group picks one up and the molecule carries an overall charge of +1. Near pH 6 the molecule exists mainly as the zwitter ion with no net charge. In strongly alkaline solution (pH ≥ 10) the \(\mathrm{-\overset{+}{N}H_3}\) group loses its proton to the base present, and the molecule carries an overall charge of −1.
At the physiological pH of 7.4, neutral α-amino acids are primarily in their zwitterionic forms. At this same pH, acidic α-amino acids exist as anions, due to deprotonation of the side-chain carboxyl group, while basic α-amino acids exist as cations, due to protonation of the side-chain amino group. The ionic structures of the constituent α-amino acids result in the ionic nature of proteins.
Glycine is dissolved separately in three solutions of pH 1, pH 6 and pH 12. Draw the structure of the form in which glycine mainly exists in each solution, and state the net charge on it in each case.
- What does the enzyme pepsin do?
- What are the initial and final products of digestion of proteins?
14.3.2 Peptide Bond and Protein
Proteins are known to break down into peptides in the stomach and duodenum under the influence of enzymes, pepsin being one of them, which is secreted by the stomach. Polypeptides are further broken down to α-amino acids. This sequence is set out in FC 14.5.
The fact that proteins break down into amino acids implies that proteins are formed by connecting α-amino acids to each other. The bond that connects α-amino acids to each other is called a peptide bond.
Consider, for example, the linking of a molecule of glycine with that of alanine. One way of doing this is to combine the carboxyl group of glycine with the α-amino group of alanine. This results in the elimination of a water molecule and the formation of a dipeptide called glycylalanine, in which the two amino acid units are linked by a peptide bond (Fig. 14.17).
\[\mathrm{H_2N-CH_2-COOH} + \mathrm{H_2N-CH(CH_3)-COOH} \xrightarrow{\ -\mathrm{H_2O}\ } \mathrm{H_2N-CH_2-CO-NH-CH(CH_3)-COOH}\]
A peptide bond is the \(\mathrm{-CO-NH-}\) linkage formed between the carboxyl group of one α-amino acid and the amino group of another, with elimination of a molecule of water.
It can be seen that a peptide bond or peptide linkage is the same as what is described as a secondary amide in organic chemistry. Nothing new has to be learnt about its chemistry; only its position in a long chain is new.
From Dipeptide to Protein
The combination of a third molecule of an α-amino acid with a dipeptide would result in the formation of a tripeptide. Similarly, linking of four, five or six α-amino acids results in the formation of a tetrapeptide, pentapeptide or hexapeptide respectively.
When the number of α-amino acids linked by peptide bonds is more than ten, the products are called polypeptides. The -CHR- units linked by peptide bonds are referred to as ‘amino acid residues’.
Proteins are polypeptides having more than a hundred amino acid residues linked by peptide bonds. It may, however, be noted that the distinction between proteins and polypeptides is not sharp.
The Two Ends of a Chain
The two ends of a polypeptide chain of a protein are not identical. The end having a free carboxyl group is called the C-terminal, while the other end having a free amino group is called the N-terminal. In the dipeptide glycylalanine, the glycine residue is N-terminal and the alanine residue is C-terminal.
This is why the order in which the two amino acids are joined matters, and why the name of a peptide is read from the N-terminal end.
- Write the structural formula of the dipeptide formed by combination of the carboxyl group of alanine and the amino group of glycine.
- Name the resulting dipeptide.
- Is this dipeptide the same as glycylalanine or its structural isomer?
Glycine, alanine and serine are available. (i) How many different tripeptides can be formed using each of the three amino acids exactly once? (ii) Write the name and the N-terminal residue of any one of them.
14.3.3 Types of Proteins
Depending upon the molecular shape, proteins are classified into two types.
a. Globular proteins: Molecules of globular proteins have a spherical shape. This shape results from the coiling around of the polypeptide chain of the protein. Globular proteins are usually soluble in water. For example: insulin, egg albumin, serum albumin, legumelin (protein in pulses).
b. Fibrous proteins: Molecules of fibrous proteins have an elongated, rod like shape. This shape is the result of holding the polypeptide chains of the protein parallel to each other. Hydrogen bonds and disulfide bonds are responsible for this shape. Fibrous proteins are insoluble in water. For example: keratin (present in hair, nail, wool), myosin (protein of muscles).
Globular proteins vs Fibrous proteins
| Molecular shape |
Spherical |
Elongated, rod like |
| How the shape arises |
Coiling around of the polypeptide chain |
Polypeptide chains held parallel to each other |
| Forces chiefly responsible |
Folding of a single chain |
Hydrogen bonds and disulfide bonds between chains |
| Solubility in water |
Usually soluble |
Insoluble |
| Examples |
Insulin, egg albumin, serum albumin, legumelin |
Keratin (hair, nail, wool), myosin (muscle) |
The shapes of protein molecules are the result of the four level structure of proteins, which is taken up next.
14.3.4 Structure of Proteins
Proteins are responsible for a variety of functions in organisms. Proteins of hair, muscles and skin give shape to the structure, while enzymes are proteins which catalyze physiological reactions.
These diverse functions of proteins can be understood by studying the four level structure of proteins, namely primary, secondary, tertiary and quaternary structure of proteins. The four levels are summarised in FC 14.6.
a. Primary Structure of Proteins
The primary structure of proteins is the sequence of constituent α-amino acid residues linked by peptide bonds. Any change in the sequence of amino acid residues results in a different protein.
The primary structure of a protein is the sequence in which its constituent α-amino acid residues are linked by peptide bonds.
The primary structure of proteins is represented by writing the three letter symbols of the amino acid residues as per their sequence in the concerned protein. The symbols are separated by dashes.
According to the convention, the N-terminal amino acid residue is written at the left end and the C-terminal amino acid residue at the right end (Fig. 14.18).
chymotrypsin (Chymotrypsin is a digestive enzyme that hydrolyses amide (peptide) bonds → it acts only where the carbonyl group of the peptide bond comes from phenylalanine, tyrosine or tryptophan → because the cut is made at fixed, known residues, chymotrypsin is used to break a long chain into a small set of predictable fragments, which is how the sequence of a protein can be worked out piece by piece.)
Chymotrypsin is a digestive enzyme that hydrolyzes those amide bonds for which the carbonyl group comes from phenylalanine, tyrosine or tryptophan. Write the symbols of the amino acids and peptides smaller than a pentapeptide formed by hydrolysis of the following hexapeptide with chymotrypsin.
Gly-Tyr-Gly-Ala-Phe-Val
In the given hexapeptide, hydrolysis by chymotrypsin can take place at two points, namely Phe and Tyr. The carbonyl group of these residues is towards the right side, that is, toward the C-terminal. Therefore the hydrolysis products in the required range will be:
Gly-Tyr, Gly-Ala-Phe and Val
(a dipeptide) (a tripeptide) (α-amino acid)
Write down the structures of amino acids constituting the following peptide.
\[\mathrm{CH_3-CH(NH_2)-CO-NH-CH(CH_2OH)-CO-NH-CH(CH_2SH)-COOH}\]
The given peptide has two amide bonds linking three amino acids. The structures of these amino acids are obtained by adding one \(\mathrm{H_2O}\) molecule across each amide bond — the -OH goes to the carbonyl carbon and the -H to the nitrogen, which is exactly the reverse of the water elimination that formed the bond.
\[\mathrm{CH_3-CH(NH_2)-COOH} + \mathrm{H_2N-CH(CH_2SH)-COOH} + \mathrm{H_2N-CH(CH_2OH)-COOH}\]
A tripeptide gives glycine, alanine and cysteine on complete hydrolysis. Write all the possible sequences of this tripeptide using three letter symbols, marking the N-terminal residue in each.
b. Secondary Structure of Proteins
The three-dimensional arrangement of localized regions of a protein chain is called the secondary structure of protein.
Hydrogen bonding between the N-H proton of one amide linkage and the C=O oxygen of another gives rise to the secondary structure. Both of these groups are already present in every peptide bond, so no special side chain is needed for this level of structure.
The secondary structure of a protein is the three-dimensional arrangement of localized regions of the protein chain, produced by hydrogen bonding between the N-H proton of one amide linkage and the C=O oxygen of another.
Two types of secondary structures commonly found in proteins are the α-helix and the β-pleated sheet.
α-Helix: The α-helix forms when a polypeptide chain twists into a right handed or clockwise spiral (Fig. 14.19). Some characteristic features of the α-helical structure of protein are:
- Each turn of the helix has 3.6 amino acids.
- A C=O group of one amino acid is hydrogen bonded to the N-H group of the fourth amino acid along the chain.
- Hydrogen bonds are parallel to the axis of the helix, while R groups extend outward from the helix core.
Myosin in muscle and α-keratin in hair are proteins with almost the entire secondary structure α-helical. Not every helix in nature is of this kind: in collagen, the protein of connective tissue, the polypeptide chains have an unusual left-handed helix structure, and three strands of these chains wind around each other in a right-handed triple helix.
β-Pleated sheet: The secondary structure is called a β-pleated sheet when two or more polypeptide chains, called strands, line up side-by-side (Fig. 14.20). The β-pleated sheet structure of protein consists of extended strands of polypeptide chains held together by hydrogen bonding.
The characteristics of the β-pleated sheet structure are:
- The C=O and N-H bonds lie in the plane of the sheet.
- Hydrogen bonding occurs between the N-H and C=O groups of nearby amino acid residues in the neighbouring chains.
- The R groups are oriented above and below the plane of the sheet.
The β-pleated sheet arrangement is favoured by amino acids with small R groups. Large side chains would collide with each other above and below the sheet, since the strands lie so close together.
α-Helix vs β-Pleated sheet
| Basic shape |
A single chain twisted into a right handed (clockwise) spiral |
Two or more extended strands lined up side-by-side in a pleated sheet |
| Hydrogen bonding is between |
C=O of one amino acid and N-H of the fourth amino acid along the same chain |
N-H and C=O groups of nearby residues in neighbouring chains |
| Direction of hydrogen bonds |
Parallel to the axis of the helix |
Across, between the strands |
| Position of R groups |
Extend outward from the helix core |
Oriented above and below the plane of the sheet |
| Favoured by |
— |
Amino acids with small R groups |
| Examples |
Myosin in muscle, α-keratin in hair |
— |
Most proteins have regions of α-helix and β-pleated sheet, in addition to other random regions that cannot be characterised by either of these secondary structures. For example, spider dragline silk protein is strong due to its β-pleated sheet region, yet elastic due to the α-helical regions in it.
c. Tertiary Structure of Proteins
The three-dimensional shape adopted by the entire polypeptide chain of a protein is called its tertiary structure.
It is the result of folding of the chain in a particular manner such that the structure is itself stabilized and also has attractive interaction with the aqueous environment of the cell. The globular and fibrous proteins represent the two major molecular shapes resulting from the tertiary structure.
The tertiary structure of a protein is the three-dimensional shape adopted by the entire polypeptide chain, resulting from folding of the chain in a way that stabilizes the structure and allows favourable interaction with the aqueous surroundings of the cell.
The forces that stabilize a particular tertiary structure include hydrogen bonding, dipole-dipole attraction (due to polar bonds in the side chains), electrostatic attraction (due to the ionic groups like \(\mathrm{-COO^{-}}\) and \(\mathrm{-\overset{+}{N}H_3}\) in the side chain) and also London dispersion forces.
Finally, disulfide bonds, formed by oxidation of nearby -SH groups (in cysteine residues), are the covalent bonds which stabilize the tertiary structure (Fig. 14.21). Note the difference in strength here: the first four forces are weak and non-covalent, whereas the disulfide bond is a genuine covalent link and is much harder to break.
A protein chain has the following amino acid residues. Show and label the interactions that can be present in various pairs from these, giving rise to the tertiary level structure of protein.
\[\mathrm{-HN-CH(CH_2OH)-CO-},\quad \mathrm{-HN-CH(CH_2-Ph)-CO-},\] \[\mathrm{-HN-CH(CH_2-CO-NH_2)-CO-},\quad \mathrm{-HN-CH(CH_2-CH-Me_2)-CO-}\]
d. Quaternary Structure of Proteins
When two or more polypeptide chains with folded tertiary structures come together into one protein complex, the resulting shape is called the quaternary structure of the protein. Each individual polypeptide chain is called a subunit of the overall protein.
For example, haemoglobin consists of four subunits called haeme, held together by intermolecular forces in a compact three dimensional shape. Haemoglobin can do its function of oxygen transport only when all the four subunits are together.
Figure 14.22 summarizes schematically the four levels of protein structure.
State the level of protein structure — primary, secondary, tertiary or quaternary — that each of the following describes, and give a one-line reason: (i) the α-helix in α-keratin, (ii) the sequence Ala-Gly-Ser-Tyr, (iii) the four haeme subunits of haemoglobin acting together, (iv) a disulfide bond folding one long chain into a compact globule.
What is the physical change observed when (a) an egg is boiled, (b) milk gets curdled on adding lemon juice?
14.3.5 Denaturation of Proteins
High temperature, acid, base and even agitation can disrupt the noncovalent interactions responsible for a specific shape of a protein. This is denaturation of protein.
Denaturation is the process by which the molecular shape of a protein changes without breaking the amide/peptide bonds that form the primary structure.
The wording of that definition is exact and worth reading twice. The weak forces — hydrogen bonds, dipole-dipole and electrostatic attractions, dispersion forces — are the ones that give way. The covalent peptide bonds of the primary structure survive, so the sequence of amino acid residues is unchanged.
Denaturation results in disturbing the secondary, tertiary or quaternary structure of the protein. This causes a change in the properties of the protein, and the biological activity is often lost.
Globular proteins are typically folded with hydrophobic side chains in the interior and polar residues on the outside, and thereby are water soluble. Denaturation exposes the hydrophobic region of globular proteins and makes them water insoluble — which is exactly what is seen when a clear egg white turns into a solid white mass. The unfolding is drawn in Fig. 14.34.
Q3. When an egg is boiled, no peptide bond in its protein is broken — yet the change is permanent and the egg white never becomes clear liquid again. What exactly has changed in the protein?
A. Boiling supplies enough energy to disrupt the weak, noncovalent interactions — hydrogen bonds, dipole-dipole and electrostatic attractions and dispersion forces — that were holding the egg albumin in its folded globular shape. The secondary, tertiary and quaternary structure collapses, while the peptide bonds of the primary structure stay intact. The chains, once folded with their hydrophobic side chains tucked inside, now lie open with those regions exposed, so the protein becomes water insoluble and sets as an opaque white solid. Since the folding was never a chemical bond that can simply be re-formed on cooling, the change does not reverse.
Denaturation is used far more often than it is named. Curdling milk with lemon juice is acid denaturation of milk protein, and it is the first step in making paneer. Boiling drinking water and sterilising surgical instruments with heat both work by denaturing the proteins of microorganisms so that their enzymes stop functioning. Even the swab of alcohol before an injection acts the same way on bacteria at the skin surface.
14.3.6 Enzymes
- Which parameter, equilibrium constant or activation energy, decides the rate of a chemical reaction?
- What is the influence of a catalyst on activation energy?
A very large number of chemical reactions take place in our bodies. These are brought about at the physiological pH of 7.4 and the body temperature of 37 °C with the help of biological catalysts called enzymes.
Those are mild conditions. The same reactions carried out in a laboratory would often need strong acids, high temperatures or high pressures — which is a measure of how effective enzymes are.
For example: insulin, a hormone secreted by the pancreas, controls blood sugar level; amylase, an enzyme present in saliva, hydrolyzes starch.
Chemically, enzymes are proteins. Every living cell contains at least 1000 different enzymes.
Enzyme Catalysis Is Highly Specific
Most enzymes catalyse only one reaction or one group of similar reactions. Thus, enzyme catalysis is highly specific.
You have learnt that a mineral acid can catalyse hydrolysis of many types of compounds such as esters, acetals and amides. In contrast, an enzyme that catalyses hydrolysis of an amide will not work on an ester or an acetal.
Mineral acid catalyst vs Enzyme catalyst
| Chemical nature |
Simple inorganic acid |
Protein |
| Range of substrates |
Catalyses hydrolysis of many types of compounds — esters, acetals, amides |
Usually one reaction or one group of similar reactions only |
| Conditions needed |
Often high temperature and concentrated acid |
Physiological pH 7.4 and body temperature 37 °C |
| Specificity |
Low |
Very high |
Enzymes are proteins that act as biological catalysts, speeding up specific reactions in living organisms under mild conditions of pH and temperature.
Mechanism of Enzyme Catalysis
The action of an enzyme on a substrate is described as the lock-and-key mechanism (Fig. 14.23). Accordingly, an enzyme has an active site on its surface. A substrate molecule can attach to this active site only if it has the right size and shape — just as only one key fits a given lock, which is where the name comes from.
Once in the active site, the substrate is held in the correct orientation to react and forms the products of the reaction. The products then leave the active site, and the enzyme is then ready to act as a catalyst again. The full cycle is set out in FC 14.7.
Formation of the enzyme-substrate complex has very low activation energy. That is how the rate of the reaction is very high. Some enzymes are so efficient that one enzyme molecule can catalyse the reaction of 10000 substrate molecules in one second.
Enzymes at Industrial Scale
Several enzymes have been isolated from organisms (such as bacteria), purified and crystallised, and the amino acid sequences of many of them have been determined.
In many industrial processes, specific reactions are carried out by use of enzymes extracted from organisms, and also by use of new enzymes made using genetic engineering.
Some examples of industrial application of enzyme catalysis are:
- Conversion of glucose to sweet-tasting fructose, using glucose isomerase.
- Manufacture of new antibiotics, using penicillin G acylase.
- Manufacture of laundry detergents, using proteases.
- Manufacture of esters used in cosmetics, using a genetically engineered enzyme.
The protease in a laundry detergent is doing exactly what pepsin does in the stomach — hydrolysing the peptide bonds of protein stains such as blood, egg and milk. There is a green-chemistry gain in this that is easy to miss: because the enzyme works at 30–40 °C, an enzyme detergent removes such stains in cool water, where an ordinary detergent would need a hot wash. Most of the energy used by a washing machine goes into heating water, so replacing harsh chemical action with a biological catalyst cuts energy use for every wash.
A patient’s stomach enzyme pepsin stops working after the person drinks a strongly alkaline solution, and does not recover even when the pH is brought back to normal. (i) Name the process that has occurred. (ii) State which levels of protein structure have been disturbed and which one has not. (iii) Explain why the loss of shape means the loss of catalytic activity.
14.4 Nucleic Acids
What is the single term that answers all the following questions?
- What decides whether you are blue eyed or brown eyed?
- Why does a wheat grain germinate to produce a wheat plant and not a rice plant?
- Which acid molecules are present in the nuclei of living cells?
One of the most remarkable properties of living cells is their ability to produce their replicas through thousands of generations. This becomes possible because a certain type of information is passed unchanged from one generation to the next. Such information is called genetic information, and its transfer to new cells is accomplished by nucleic acids.
There are two types of nucleic acids: ribonucleic acids (RNA) and deoxyribonucleic acids (DNA). RNA are found mainly in the fluid of living cells (cytoplasm), while DNA are found primarily in the nuclei of living cells. The word “nucleic” comes from exactly this observation — these acids were first obtained from cell nuclei.
14.4.1 Nucleotides
Nucleic acids are unbranched polymers of repeating monomers called nucleotides. In other words, nucleic acids have a polynucleotide structure.
The scale is very different for the two kinds. DNA molecules contain several million nucleotides, while RNA molecules contain a few thousand nucleotides.
The nucleotide monomers consist of three components: a monosaccharide, a nitrogen-containing base and a phosphate group.
A nucleotide is the repeating monomer unit of a nucleic acid, made up of a monosaccharide, a nitrogen-containing base and a phosphate group.
The Sugar Component
Nucleotides of both RNA and DNA contain a five-membered ring monosaccharide (a furanose), often called simply the sugar component. You met the furanose ring earlier in fructose (section 14.2.8); the same five-membered ring appears here.
In RNA, the sugar component of the nucleotide unit is D-ribose, while in DNA it is 2-deoxy-D-ribose (Fig. 14.24). The prefix “deoxy” means exactly what it says: this sugar is ribose with one oxygen missing, since there is no -OH group at C-2.
The Nitrogen-Containing Bases
In total, five nitrogen-containing bases are present in nucleic acids.
Three bases with one ring — cytosine, uracil and thymine — are derived from the parent compound pyrimidine. Two bases with two rings — adenine and guanine — are derived from the parent compound purine. Each base is designated by a one-letter symbol (Fig. 14.25).
Uracil (U) occurs only in RNA, while thymine (T) occurs only in DNA. Adenine, guanine and cytosine are common to both.
DNA vs RNA — components
| Sugar |
D-2-deoxyribose |
D-ribose |
| Purine bases |
A, G |
A, G |
| Pyrimidine bases |
C, T |
C, U |
| Base found only here |
Thymine (T) |
Uracil (U) |
| Where found in the cell |
Primarily in the nuclei |
Mainly in the cytoplasm |
| Number of nucleotides per molecule |
Several million |
A few thousand |
A nucleic acid on complete hydrolysis is found to give D-ribose, phosphoric acid and the bases adenine, guanine, cytosine and uracil. (i) Identify the nucleic acid. (ii) Name the base whose absence helped you decide. (iii) State whether this nucleic acid is single stranded or double stranded.
From Nucleoside to Nucleotide
A nucleoside is formed by joining the anomeric carbon of the furanose with the nitrogen of a base. Recall from section 14.2.6 that the anomeric carbon is the one that was the carbonyl carbon before the ring closed; in the sugars of nucleic acids it is C-1.
While numbering the atoms in a nucleoside, primes (’) are used for the furanose numbering, to distinguish them from the atoms of the base (Fig. 14.26). So “C-1” belongs to the base and “C-1′” belongs to the sugar — a small convention that prevents a great deal of confusion.
With pyrimidine bases, the nitrogen atom at the 1 position bonds with the 1′ carbon of the sugar. With purine bases, the nitrogen atom at the 9 position bonds with the 1′ carbon of the sugar.
Nucleotides are formed by adding a phosphate group to the 5′-OH of a nucleoside (Fig. 14.27). Thus, nucleotides are monophosphates of nucleosides.
Abridged names of some nucleotides are AMP, dAMP, UMP, dTMP and so on. These are read as follows: the first capital letter is derived from the corresponding base, MP stands for monophosphate, and the small letter ‘d’ at the beginning indicates deoxyribose in the nucleotide.
Name the nucleotide that contains 2-deoxy-D-ribose, the base thymine and one phosphate group, and state whether it belongs to DNA or to RNA.
Build the abridged name from its three parts, in the order given by the convention above.
The base is thymine, so the capital letter is T. One phosphate group is present, so the ending is MP, standing for monophosphate. The sugar is 2-deoxy-D-ribose, so the name carries a small d at the beginning.
\[\text{Abridged name} = \mathbf{dTMP}\]
Thymine occurs only in DNA, and the sugar 2-deoxy-D-ribose also occurs only in DNA. Both pieces of evidence agree, so dTMP is a nucleotide of DNA.
Draw structural formulae of nucleosides formed from the following sugars and bases.
- D-ribose and guanine
- D-2-deoxyribose and thymine
Write the abridged name of the nucleotide made from D-ribose, uracil and one phosphate group. State one reason why this nucleotide can never occur in DNA.
14.4.2 Structure of Nucleic Acids
Nucleic acids, both DNA and RNA, are polymers of nucleotides, formed by joining the 3′-OH group of one nucleotide with the 5′-phosphate of another nucleotide (Fig. 14.28). The linkage so formed, in which one phosphate group is bonded to two sugar units, is called a phosphodiester linkage.
The two ends of a polynucleotide chain are distinct from each other. One end, having the free phosphate group at the 5′ position, is called the 5′ end. The other end is the 3′ end and has a free OH group at the 3′ position.
Because the two ends are different, a polynucleotide chain has a direction, in the same way that a polypeptide chain has an N-terminal and a C-terminal end (section 14.3.2).
The Sugar-Phosphate Backbone
A nucleic acid contains a backbone consisting of alternating sugar and phosphate groups, with the bases hanging off the sugars. The backbone of all types of DNA contains the sugar 2-deoxy-D-ribose, while that of RNA contains the sugar D-ribose.
The polynucleotide structure of nucleic acids can be represented schematically as in Fig. 14.29 (a and b).
Primary Structure of Nucleic Acids
The primary structure of nucleic acids is the sequence of the nucleotides in it. This, in turn, is determined by the identity of the bases in the nucleotides, since the sugar and the phosphate repeat unchanged all along the backbone.
Different nucleic acids have distinct primary structures. It is the sequence of bases in DNA which carries the genetic information of the organism.
The identity and sequence of bases therefore distinguish one polynucleotide from another. A polynucleotide has one free phosphate group at the 5′ end and a free OH group at the 3′ end.
The polynucleotide chains of nucleic acids are named by the sequence of the bases, beginning at the 5′ end and using the one letter symbols of the bases. For example, the name CATG means there are four nucleotides in the segment, containing the bases cytosine, adenine, thymine and guanine, in the indicated order from the 5′ end.
Q5. DNA is built from only four different bases, yet it stores the instructions for making thousands of different proteins in one organism. How can so few building blocks hold so much information?
A. The information is not in the bases themselves but in their order. Each position along the chain can be filled by any one of four bases, so a segment of just ten nucleotides already has \(4^{10}\), that is more than a million, possible sequences — and a DNA molecule contains several million nucleotides. Compare this with the twenty-six letters of the alphabet, which are enough for every book ever written, because meaning lies in the sequence and not in the letters. Since a gene is simply a portion of the DNA whose base sequence specifies one protein, a single DNA molecule has room for thousands of such genes.
Draw a schematic representation of the trinucleotide segment ‘ACT’ of a DNA molecule.
In a DNA molecule the sugar is deoxyribose. The base ‘A’ in the given segment is at the 5′ end while the base ‘T’ is at the 3′ end, because the sequence is always read from the 5′ end.
Hence the schematic representation of the given segment of DNA is a chain of alternating boxes: Phosphate–deoxyribose–Phosphate–deoxyribose–Phosphate–deoxyribose, with the 5′ end at the left and the 3′ end at the right, and with the base A hanging below the first deoxyribose, C below the second and T below the third.
Draw a schematic representation of the trinucleotide segment ‘GUC’ of an RNA molecule, labelling the 5′ and 3′ ends. State two ways in which your diagram differs from the schematic representation of a DNA segment.
14.4.3 DNA Double Helix
James Watson and Francis Crick put forth in 1953 a double helix model for DNA structure, which was later verified by electron microscopy. The salient features of the Watson and Crick model of DNA are:
- DNA consists of two polynucleotide strands that wind into a right-handed double helix.
- The two strands run in opposite directions; one from the 5′ end to the 3′ end, while the other runs from the 3′ end to the 5′ end.
- The sugar-phosphate backbone lies on the outside of the helix and the bases lie on the inside, perpendicular to the axis of the helix.
- The double helix is stabilized by hydrogen bonding between the bases of the two DNA strands. This gives rise to a ladder-like structure of the DNA double helix, in which the two backbones are the side rails and the paired bases are the rungs.
- Adenine always forms two hydrogen bonds with thymine, and guanine forms three hydrogen bonds with cytosine. Thus A-T and C-G are complementary base pairs, and the two strands of the double helix are complementary to each other.
The pairing is not a matter of chance. A purine (two rings) always pairs with a pyrimidine (one ring), so that every rung of the ladder has the same width, and the positions of the -NH and C=O groups on each pair line up exactly for hydrogen bonding. As shown in Fig. 14.33, no other combination of bases fits.
It may be noted that RNA exists as a single stranded structure.
DNA vs RNA — overall structure
| Number of strands |
Two, wound into a right-handed double helix |
Single stranded |
| Base pairing present |
Yes — A-T and C-G |
No pairing between two strands |
| Function |
Stores the genetic information |
Translates the information into protein synthesis |
| Location |
Primarily in the nuclei of cells |
Mainly in the cytoplasm |
The 1953 double helix model was not built by making a new compound but by fitting together evidence that already existed — the chemistry of the bases, the known base ratios, and above all the X-ray diffraction photographs of DNA fibres obtained at King’s College London by Rosalind Franklin, whose sharpest image showed the tell-tale pattern of a helix. Watson and Crick assembled these clues into a physical model, and the structure immediately suggested how genetic information could be copied: separate the two complementary strands, and each one is a template for rebuilding the other.
Write the sequence of the complementary strand of the following portion of a DNA molecule: 5′ - ACGTAC - 3′
The complementary strand runs in the opposite direction, from the 3′ end to the 5′ end. It has the base sequence decided by the complementary base pairs A - T and C - G.
Original strand: 5′ - A C G T A C - 3′
Complementary strand: 3′ - T G C A T G - 5′
Each base of the original strand is simply replaced by its partner: A by T, T by A, C by G and G by C, position by position.
A segment of one DNA strand has the sequence 5′ - GCATG - 3′. Calculate the total number of hydrogen bonds holding this segment to its complementary strand.
First write the base pairs, taking each base of the given segment with its partner.
G-C, C-G, A-T, T-A, G-C
Now apply the rule that a G-C pair is held by 3 hydrogen bonds and an A-T pair by 2 hydrogen bonds.
Number of G-C pairs = 3, number of A-T pairs = 2.
\[\text{Total hydrogen bonds} = (3 \times 3) + (2 \times 2) = 9 + 4 = 13\]
So 13 hydrogen bonds hold the two segments together.
base composition of a DNA sample (A sample of DNA is completely hydrolysed and the four bases are estimated. The percentages found are shown in Fig. 14.35. Answer the following from the data: (i) Which two pairs of bases occur in almost equal amounts, and why should this be so? (ii) What would you expect the percentage of cytosine to be in a DNA sample whose guanine content is 22%? (iii) A second sample is found to contain uracil instead of thymine — what does this tell you about it, and would you still expect the paired equality you found in part (i)? Give reasons for each answer.)
Complementary base pairing is not only a structural idea; it is the basis of a working technology. In DNA fingerprinting, a sample from a crime scene or a disputed parentage case is compared with reference samples by matching the base sequences in selected regions of the DNA, and the match is read out through short synthetic strands that bind only to their exact complement. The same principle underlies the diagnostic tests that detect a virus by searching for a specific stretch of its genetic material in a patient’s sample. In each case the chemistry being used is simply the rule that A pairs with T and C pairs with G.
A double stranded DNA molecule contains 1200 base pairs, of which 400 are adenine-thymine pairs. (i) How many guanine-cytosine pairs are present? (ii) Calculate the total number of hydrogen bonds holding the two strands together. (iii) State which of the two strands, if either, contains more adenine — and justify your answer.
14.5 Summary
Biomolecules are the molecules that build the structure of living organisms and take part in the physiological processes inside them. Proteins and cellulose are the primary structural materials, plants make carbohydrates by photosynthesis and build proteins from minerals absorbed by their roots, lipids are the main ingredient of vegetable oils and milk fats, and nucleic acids are the genetic material. This chapter studies three of these classes — carbohydrates, proteins and nucleic acids.
Carbohydrates are polyhydroxy aldehydes or polyhydroxy ketones, or compounds which give such units on hydrolysis. They are also called saccharides, from the Greek sakkharon, and the name carbohydrate comes from the fact that many of their formulae can be written as \(\mathrm{C}_x(\mathrm{H_2O})_y\) — for example glucose as \(\mathrm{C_6(H_2O)_6}\) and starch as \(\mathrm{[C_6(H_2O)_5]_n}\).
On the basis of their behaviour on hydrolysis, carbohydrates are of three kinds. Monosaccharides do not hydrolyse further. Oligosaccharides give two to ten monosaccharide units and are further classed as disaccharides, trisaccharides and so on. Polysaccharides give a very large number of units on complete hydrolysis. The overall scheme is condensed in FC 14.10.
A monosaccharide is generally named glycose. One with an aldehydic carbonyl group is an aldose, one with a ketonic carbonyl group is a ketose, and the name is completed by the number of carbons. Glucose is an aldohexose and fructose is a ketohexose, both being \(\mathrm{C_6H_{12}O_6}\).
Glucose is prepared by acid catalysed hydrolysis of sucrose, using dilute hydrochloric or sulfuric acid for about two hours, glucose being separated by adding ethanol during cooling; and commercially from starch by boiling with dilute sulfuric acid at 393 K under 2 to 3 atm pressure.
\[\mathrm{C_{12}H_{22}O_{11}} + \mathrm{H_2O} \xrightarrow{\ \mathrm{H^{\oplus}},\ \Delta\ } \mathrm{C_6H_{12}O_6} + \mathrm{C_6H_{12}O_6}\]
\[\mathrm{(C_6H_{10}O_5)_n} + n\,\mathrm{H_2O} \xrightarrow{\ \mathrm{H^{\oplus}},\ 393\ \text{K},\ 2\text{--}3\ \text{atm}\ } n\,\mathrm{C_6H_{12}O_6}\]
- The open chain structure of glucose, \(\mathrm{CHO\!-\!(CHOH)_4\!-\!CH_2OH}\), rests on six pieces of chemical evidence: its molecular formula \(\mathrm{C_6H_{12}O_6}\); formation of n-hexane with HI, showing a straight chain of six carbons; formation of an oxime and a cyanohydrin, showing one carbonyl group; oxidation by bromine water to gluconic acid, showing that the carbonyl group is an aldehyde; formation of glucose pentaacetate with five moles of acetic anhydride, showing five hydroxyl groups on five different carbons; and formation of the same saccharic acid from both glucose and gluconic acid with dilute nitric acid, showing one primary alcoholic group.
\[\mathrm{CHO\!-\!(CHOH)_4\!-\!CH_2OH} \xrightarrow{\ [\mathrm{O}],\ \mathrm{Br_2}\ \text{water}\ } \mathrm{COOH\!-\!(CHOH)_4\!-\!CH_2OH}\]
\[\mathrm{CHO\!-\!(CHOH)_4\!-\!CH_2OH} \xrightarrow{\ [\mathrm{O}],\ \mathrm{HNO_3}\ } \mathrm{COOH\!-\!(CHOH)_4\!-\!COOH}\]
Glucose contains four chiral carbons, and a molecule with \(n\) chiral carbons can have a maximum of \(2^n\) optical isomers; an aldohexose therefore has \(2^4 = 16\) optical isomers. Glucose has specific rotation \([\alpha]_D^{20} = +52.7^{\circ}\) and is called dextrose.
The prefixes D- and L- give the relative configuration of a stereoisomer with reference to glyceraldehyde, and are assigned from the lowest chiral carbon in the Fischer projection formula. The capital letter describes configuration only, not the direction in which the compound rotates plane-polarised light.
Glucose also exists in ring form, produced by reaction of the formyl group with the \(\mathrm{-OH}\) group at C-5, giving a hemiacetal. The two ring forms differ only at C-1, the anomeric carbon, and are called the \(\alpha\)- and \(\beta\)-anomers; both are in equilibrium with the open chain form in aqueous solution. The six-membered ring is a pyranose ring, so glucose is also called glucopyranose, and the ring is best drawn as a Haworth formula, in which the lower side is the \(\alpha\)-side and the upper side the \(\beta\)-side.
The hemiacetal group of glucopyranose is a potential aldehyde group, which is why glucose gives positive Tollens and Fehling tests and is a reducing sugar.
Fructose is a laevorotatory ketohexose with \([\alpha]_D^{20} = -92.4^{\circ}\), also called laevulose. Being an \(\alpha\)-hydroxy keto compound it too is a reducing sugar. In the free state it is mainly fructopyranose, while in the combined state, as in sucrose, it is fructofuranose; its ring is a hemiketal.
In disaccharides two monosaccharide units are joined by a glycosidic linkage, an \(\mathrm{-O-}\) bridge formed by loss of a water molecule between two hydroxyl groups, at least one of which must be an anomeric hydroxyl group. Sucrose (\(+66.5^{\circ}\)) has a linkage between C-1 of \(\alpha\)-glucose and C-2 of \(\beta\)-fructose, uses both anomeric groups, and is non-reducing; its hydrolysis with dilute acid or invertase gives equimolar glucose and fructose, has net laevorotation, and is therefore called inversion, the product being invert sugar. Maltose is two D-glucose units joined by an \(\alpha\)-1,4-glycosidic linkage and is reducing, giving glucose with dilute acid or maltase. Lactose, the sugar of milk, is D-galactose joined to D-glucose by a \(\beta\)-1,4-glycosidic linkage and is also reducing.
Polysaccharides are large numbers of monosaccharide units joined by glycosidic linkages. Starch, the storage carbohydrate of plants, is a polymer of \(\alpha\)-D-glucose with two components: amylose (15-20%), water soluble, unbranched, 200-1000 \(\alpha\)-glucose units joined by \(\alpha\)-1,4 linkages, giving a blue colour with iodine; and amylopectin (80-85%), water insoluble and branched, with \(\alpha\)-1,4 linkages in the chains and \(\alpha\)-1,6 linkages at the branches, giving a blue-violet colour with iodine. Cellulose is a straight chain of \(\beta\)-glucose units joined by \(\beta\)-1,4 linkages, is the main constituent of plant and bacterial cell walls, wood and cotton, and cannot be digested by humans, serving as dietary fibre. Glycogen resembles amylopectin but is more highly branched, and is the storage carbohydrate of animals, present in liver, muscles and brain and also in yeast and fungi.
Proteins are polyamides — high molecular weight polymers of \(\alpha\)-amino acid monomers. The name comes from the Greek proteios, meaning primary. Milk, pulses, nuts, fish and meat are nutritional sources.
An \(\alpha\)-amino acid has an amino group on the carbon next to the carboxyl group. Its \(\alpha\)-carbon is chiral in every case except glycine, and in amino acids obtained from proteins it has the L configuration. The side chain R decides the type: a carboxyl group in R makes the amino acid acidic, an amino group makes it basic, and the rest are neutral. Twenty amino acids are commonly found in proteins, written as three letter or one letter symbols, and ten of them are essential amino acids that the human body cannot synthesise and must obtain from the diet.
Because an \(\alpha\)-amino acid carries both an acidic and a basic group, proton transfer within the molecule gives a dipolar zwitter ion. This is why \(\alpha\)-amino acids are high melting, water soluble crystalline solids unlike comparable amines and carboxylic acids. The form present depends on pH — an overall \(+1\) charge at low pH, the zwitter ion near neutral pH, and an overall \(-1\) charge at high pH — and at the physiological pH of 7.4 neutral amino acids exist mainly as zwitter ions.
The peptide bond, \(\mathrm{-CO\!-\!NH-}\), is the same as a secondary amide and joins amino acids with elimination of water; glycine and alanine so joined give the dipeptide glycylalanine. Linking more units gives tri-, tetra-, penta- and hexapeptides; more than ten gives a polypeptide, and more than a hundred residues gives a protein. The end with a free amino group is the N-terminal and the end with a free carboxyl group is the C-terminal.
\[\mathrm{H_2N\!-\!CH_2\!-\!COOH} + \mathrm{H_2N\!-\!CH(CH_3)\!-\!COOH} \xrightarrow{\ -\mathrm{H_2O}\ } \mathrm{H_2N\!-\!CH_2\!-\!CO\!-\!NH\!-\!CH(CH_3)\!-\!COOH}\]
By molecular shape, proteins are globular — spherical, formed by coiling of the chain, usually water soluble, as in insulin, egg albumin, serum albumin and legumelin; or fibrous — rod like, formed by parallel chains held by hydrogen and disulfide bonds, insoluble in water, as in keratin and myosin.
Protein structure is described at four levels, condensed in FC 14.11.
Denaturation is the process by which the molecular shape of a protein changes without breaking the amide or peptide bonds of the primary structure. High temperature, acid, base and even agitation disrupt the noncovalent interactions, the secondary, tertiary or quaternary structure is disturbed, the properties change and the biological activity is often lost.
Enzymes are proteins that act as biological catalysts at the physiological pH of 7.4 and body temperature of \(37\ ^{\circ}\mathrm{C}\), and every living cell contains at least 1000 of them. Enzyme catalysis is highly specific: unlike a mineral acid, which catalyses hydrolysis of esters, acetals and amides alike, an enzyme for amide hydrolysis will not act on an ester or acetal.
The action of an enzyme is described by the lock-and-key mechanism. A substrate of the right size and shape attaches to the active site, is held in the correct orientation and reacts; the products leave and the enzyme is free again. The enzyme-substrate complex has very low activation energy, which is why the rate is so high, and one molecule of some enzymes can catalyse the reaction of 10000 substrate molecules in one second. Industrial uses include glucose isomerase for converting glucose to fructose, penicillin G acylase for new antibiotics, proteases in laundry detergents, and genetically engineered enzymes for cosmetic esters.
Nucleic acids carry genetic information from one generation to the next. There are two kinds: ribonucleic acids (RNA), found mainly in the cytoplasm, and deoxyribonucleic acids (DNA), found primarily in the nuclei of cells. DNA stores the genetic information while RNA translates it into the synthesis of proteins, and a gene is the portion of a DNA molecule responsible for the synthesis of a single protein.
Nucleic acids are unbranched polymers of nucleotides, each nucleotide being made of a monosaccharide, a nitrogen-containing base and a phosphate group. The sugar is a five-membered furanose ring: D-ribose in RNA and 2-deoxy-D-ribose in DNA. Five bases occur — cytosine, uracil and thymine derived from pyrimidine, and adenine and guanine derived from purine — with uracil only in RNA and thymine only in DNA. The build-up is condensed in FC 14.12.
A nucleoside is formed by joining the anomeric carbon of the furanose to a nitrogen of a base, primes being used for the sugar numbering; pyrimidine bases bond through N-1 and purine bases through N-9 to the 1′ carbon. Adding a phosphate group to the 5′-OH gives a nucleotide, so nucleotides are monophosphates of nucleosides, named AMP, dAMP, UMP, dTMP and so on, where MP stands for monophosphate and a small d for deoxyribose.
In a nucleic acid, the 3′-OH group of one nucleotide is joined to the 5′-phosphate of the next through a phosphodiester linkage, giving a backbone of alternating sugar and phosphate units with the bases attached to the sugars. The chain has a 5′ end carrying a free phosphate group and a 3′ end carrying a free OH group, and is named by the sequence of one letter base symbols beginning at the 5′ end. The primary structure of a nucleic acid is this sequence of nucleotides, and it is the sequence of bases in DNA that carries the genetic information of the organism. DNA molecules contain several million nucleotides and RNA molecules a few thousand.
Watson and Crick proposed the double helix model of DNA in 1953, later verified by electron microscopy. Two polynucleotide strands wind into a right-handed double helix and run in opposite directions, one from the 5′ end to the 3′ end and the other from the 3′ end to the 5′ end. The sugar-phosphate backbone lies outside and the bases lie inside, perpendicular to the axis. Hydrogen bonding between the bases of the two strands stabilizes the helix and gives it a ladder-like structure: adenine always forms two hydrogen bonds with thymine and guanine forms three with cytosine, so A-T and C-G are complementary base pairs and the two strands are complementary to each other. RNA, in contrast, exists as a single stranded structure.
14.6 Objective Questions
- \(\mathrm{CH_2OH\!-\!CO\!-\!(CHOH)_4\!-\!CH_2OH}\) is an example of
- Aldohexose
- Aldoheptose
- Ketotetrose
- Ketoheptose
- Open chain formula of glucose does not contain
- Formyl group
- Anomeric hydroxyl group
- Primary hydroxyl group
- Secondary hydroxyl group
- Which of the following does not apply to \(\mathrm{CH_2NH_2\!-\!COOH}\)
- Neutral amino acid
- L-amino acid
- Exists as zwitter ion
- Natural amino acid
- Tryptophan is called essential amino acid because
- It contains aromatic nucleus.
- It is present in all the human proteins.
- It cannot be synthesised by human body.
- It is essential constituent of enzymes.
- A disulfide link gives rise to the following structure of protein.
- Primary
- Secondary
- Tertiary
- Quaternary
- RNA has
- A - U base pairing
- P-S-P-S backbone
- double helix
- G - C base pairing
14.7 Check Your Concepts
Assertion: Cellulose cannot be digested by human beings. Reason: Cellulose is a straight chain polysaccharide of \(\beta\)-glucose units joined by \(\beta\)-1,4-glycosidic bonds. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
Lactose and sucrose are both disaccharides with the molecular formula \(\mathrm{C_{12}H_{22}O_{11}}\). Explain why only one of them gives a positive Fehling test, referring to the hydroxyl groups used in forming the glycosidic linkage.
The six-membered ring form of glucose is called a pyranose and the five-membered ring form of fructose is called a furanose. Explain the origin of both names, and state which ring form of fructose is present in sucrose.
Explain why the carbon atom C-1 of glucose is not chiral in the open chain structure but becomes chiral when the ring closes. What name is given to this carbon, and what name is given to the pair of ring structures that differ only at this position?
Glucose is described both as (+)-glucose and as D-glucose. State what each designation conveys, and explain why one of them cannot be predicted from the other.
Amylose and amylopectin are both built only from \(\alpha\)-D-glucose units, yet one is soluble in water and the other is not, and they give different colours with iodine. Account for both differences in terms of the structures of the two components.
Match the linkage in Column I with the substance in Column II in which it is found.
| a. \(\alpha\)-1,4-glycosidic linkage |
i. Cellulose |
| b. \(\beta\)-1,4-glycosidic linkage |
ii. Branch point of amylopectin |
| c. \(\alpha\)-1,6-glycosidic linkage |
iii. Sucrose |
| d. Linkage between C-1 of \(\alpha\)-glucose and C-2 of \(\beta\)-fructose |
iv. Maltose |
Raffinose and stachyose are oligosaccharides. Name the class to which each belongs and write the monosaccharide units obtained from each on complete hydrolysis.
Glucose on prolonged heating with HI gives n-hexane. Explain what this observation establishes, and state what would have been concluded had a branched hexane been obtained instead.
Glucose gives gluconic acid with bromine water but saccharic acid with dilute nitric acid. Explain how the difference between the two products reveals the nature of the two end groups of the glucose molecule.
Complete the following reactions and name the organic product in each case.
- Glucose \(+\) HCN \(\rightarrow\)
- Glucose \(+ \mathrm{NH_2OH} \rightarrow\)
- Glucose \(+\) acetic anhydride \(\rightarrow\)
Assertion: Glycine is the only \(\alpha\)-amino acid in Table 14.1 that is optically inactive. Reason: The \(\alpha\)-carbon of glycine carries two hydrogen atoms. State whether the assertion and the reason are true, and whether the reason correctly explains the assertion.
Classify each of the following \(\alpha\)-amino acids as acidic, basic or neutral, giving the feature of the side chain responsible in each case: aspartic acid, lysine, serine, glutamic acid, histidine.
Alanine is dissolved in three separate solutions of pH 2, pH 6 and pH 11. Draw the form in which alanine exists in each solution and state the net charge on the molecule in each case.
What happens if a single amino acid residue in the sequence of a protein is replaced by a different one? Name the level of protein structure directly affected, and explain why the effect need not stop at that level.
The \(\beta\)-pleated sheet arrangement is favoured by amino acids with small R groups. Give the structural reason for this preference, and state where the R groups lie with respect to the plane of the sheet.
Haemoglobin cannot transport oxygen unless all four of its subunits are together. Name the level of protein structure being described, and explain why the separated subunits fail to do the job.
During denaturation of a protein, some interactions are disrupted while others survive. State which are disrupted and which survive, name the levels of structure that are lost, and explain why the biological activity is usually lost as well.
A mineral acid catalyses the hydrolysis of esters, acetals and amides alike, but an enzyme that hydrolyses amides will not act on an ester. Name the property of enzymes shown here and explain it using the lock-and-key mechanism.
Explain why a single enzyme molecule can bring about the reaction of a very large number of substrate molecules, and state why the enzyme-substrate complex makes the reaction fast.
State which nitrogen-containing base occurs only in RNA and which occurs only in DNA. Name the sugar present in each nucleic acid and state the structural difference between the two sugars.
The two ends of a polynucleotide chain are not identical. Describe each end, explain how the linkage between successive nucleotides makes them different, and state the convention followed while naming a polynucleotide by its base sequence.
Two short DNA segments of equal length are taken; one is rich in A-T pairs and the other in C-G pairs. Predict which of the two would need more energy to separate into single strands, and justify your answer.
Arrange starch, cellulose and glycogen in increasing order of branching, and state for each whether it is a storage or a structural polysaccharide.
Give scientific reasons:
- The disaccharide sucrose gives negative Tollens test while the disaccharide maltose gives positive Tollens test.
- On complete hydrolysis DNA gives equimolar quantities of adenine and thymine.
- \(\alpha\)-Amino acids have high melting points compared to the corresponding amines or carboxylic acids of comparable molecular mass.
- Hydrolysis of sucrose is called inversion.
- On boiling egg albumin becomes opaque white.
- Some of the following statements apply to DNA only, some to RNA only and some to both. Label them accordingly.
- The polynucleotide is double stranded. (……..)
- The polynucleotide contains uracil. (……..)
- The polynucleotide contains D-ribose (……..).
- The polynucleotide contains Guanine (……..).
- Write the sequence of the complementary strand for the following segments of a DNA molecule.
- 5’ - CGTTTAAG - 3’
- 5’ - CCGGTTAATACGGC - 3’
Write the names and schematic representations of all the possible dipeptides formed from alanine, glycine and tyrosine.
Give two evidences for presence of formyl group in glucose.
Draw a neat diagram for the following:
- Haworth formula of glucopyranose
- Zwitter ion
- Haworth formula of maltose
- Secondary structure of protein
- AMP
- dAMP
- One purine base from nucleic acid
- Enzyme catalysis
14.8 Practice With Numericals
An alcoholic compound of molecular mass 92 u was acetylated, and the acetyl derivative was found to have molecular mass 218 u. Calculate the number of alcoholic (\(\mathrm{-OH}\)) groups present in the original compound.
68.4 g of sucrose is completely hydrolysed with dilute hydrochloric acid. Calculate the mass of glucose and the mass of fructose obtained. (Molar mass of sucrose \(= 342\ \text{g mol}^{-1}\); molar mass of glucose \(=\) molar mass of fructose \(= 180\ \text{g mol}^{-1}\).)
Calculate the mass of glucose obtained by the complete hydrolysis of 810 g of starch. (Take the molar mass of one \(\mathrm{C_6H_{10}O_5}\) unit as \(162\ \text{g mol}^{-1}\) and the molar mass of glucose as \(180\ \text{g mol}^{-1}\).)
17.1 g of maltose is completely hydrolysed with the enzyme maltase. Calculate the mass of glucose formed. (Molar mass of maltose \(= 342\ \text{g mol}^{-1}\); molar mass of glucose \(= 180\ \text{g mol}^{-1}\).)
Calculate the maximum number of optical isomers possible for (i) an aldopentose, (ii) an aldohexose and (iii) a ketohexose, showing in each case how you counted the chiral carbon atoms.
18.0 g of glucose is completely acetylated to glucose pentaacetate. Calculate (i) the number of moles of acetic anhydride required and (ii) the mass of glucose pentaacetate formed. (Molar mass of glucose \(= 180\ \text{g mol}^{-1}\); molar mass of glucose pentaacetate \(= 390\ \text{g mol}^{-1}\).)
An unbranched amylose molecule is built from 450 \(\alpha\)-glucose units. Calculate (i) the number of \(\alpha\)-1,4-glycosidic linkages in the molecule and (ii) the number of water molecules required for its complete hydrolysis to glucose.
Calculate the molar mass of the tripeptide glycylalanylglycine formed from two molecules of glycine and one molecule of alanine. (Molar mass of glycine \(= 75\ \text{g mol}^{-1}\); molar mass of alanine \(= 89\ \text{g mol}^{-1}\); molar mass of water \(= 18\ \text{g mol}^{-1}\).)
A single unbranched polypeptide chain contains 320 amino acid residues. Calculate (i) the number of peptide bonds present and (ii) the number of water molecules eliminated during its formation.
A double stranded DNA molecule contains 900 base pairs, of which 350 are adenine-thymine pairs. Calculate (i) the number of guanine-cytosine pairs and (ii) the total number of hydrogen bonds holding the two strands together.
In a sample of double stranded DNA, adenine accounts for 22% of the total bases. Using the complementary base pairing rules, calculate the percentage of thymine, of guanine and of cytosine in the sample.
A segment of one strand of a DNA molecule has the base sequence 5’ - TAGCCGAT - 3’. Calculate (i) the number of nucleotides in the double stranded segment formed with its complementary strand and (ii) the total number of hydrogen bonds between the two strands of this segment.